Chapter 24, Problem 24.34AP

Interpretation Introduction

(a)

Interpretation:

The products expected when D-mannose is reacted with ${\text{Ag}}^{+}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}$ is to be stated.

Concept introduction:

Tollens test is the chemical test for the identification of the presence of the aldehyde group in the compound. Aldehydes are readily oxidized by the Tollens reagent into the carboxylic acid group and itself reduced to silver and produces a silver mirror. Therefore this test is also known as the silver mirror test.

Answer to Problem 24.34AP

The product obtained when D-mannose is reacted with ${\text{Ag}}^{+}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}$ is shown below.

Explanation of Solution

The product obtained when D-mannose is reacted with ${\text{Ag}}^{+}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}$ is shown below.

Figure 1

The Tollens reagent oxidizes the aldehyde group into carboxylic acid. The D-mannose is oxidized by the ${\text{Ag}}^{+}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}$ into D-mannonic acid.

Conclusion

The product obtained when D-mannose is reacted with ${\text{Ag}}^{+}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}$ is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation:

The products expected when D-mannose is reacted with dilute $\text{HCl}$ is to be stated.

Concept introduction:

There are two forms of sugars that is $\alpha$ and $\beta$ forms. The two forms of sugars are in equilibrium with each other in the presence of the dilute acid or base solution. Both forms of sugars have a different value of optical rotation. The change of one form into another changes the optical rotation of the solution and this change is called mutarotation.

Answer to Problem 24.34AP

The product obtained when D-mannose is reacted with dilute $\text{HCl}$ is shown below.

Explanation of Solution

The product obtained when D-mannose is reacted with dilute $\text{HCl}$ is shown below.

Figure 2

The conversion of both forms of sugar into each other in the presence of acid and base. That is the alpha form is converted beta and vice-versa. This change occurs until an equilibrium mixture of both compounds is obtained.

Conclusion

The product obtained when D-mannose is reacted with dilute $\text{HCl}$ is shown in Figure 2.

Interpretation Introduction

(c)

Interpretation:

The products expected when D-mannose is reacted with dilute $\text{NaOH}$ is to be stated.

Concept introduction:

There are two forms of sugars that is $\alpha$ and $\beta$ forms. The two forms of sugars are in equilibrium with each other in the presence of the dilute acid or base solution. Both forms of sugars have a different value of optical rotation. The change of one form into another changes the optical rotation of the solution and this change is called mutarotation.

Answer to Problem 24.34AP

The product obtained when D-mannose is reacted with dilute $\text{NaOH}$ is shown below.

Explanation of Solution

The product obtained when D-mannose is reacted with dilute $\text{NaOH}$ is shown below.

Figure 3

The conversion of both forms of sugar into each other in the presence of acid or base. That is the alpha form is converted beta and vice-versa. This change occurs until an equilibrium mixture of both compounds is obtained.

Conclusion

The product obtained when D-mannose is reacted with dilute $\text{NaOH}$ is shown in Figure 3.

Interpretation Introduction

(d)

Interpretation:

The products expected when D-mannose is reacted with ${\text{Br}}_{\text{2}}/{\text{H}}_{\text{2}}\text{O}$, then ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is to be stated.

Concept introduction:

The aldehyde group on oxidation changes into a carboxylic acid. The oxidation of the aldehyde group present in the aldoses is done by the bromine and water into carboxylic acids. The carboxylic acid of the aldoses thus formed is converted into lactones form.

Answer to Problem 24.34AP

The product obtained when D-mannose is reacted with ${\text{Br}}_{\text{2}}/{\text{H}}_{\text{2}}\text{O}$, then ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is shown below.

Explanation of Solution

The product obtained when D-mannose is reacted with ${\text{Br}}_{\text{2}}/{\text{H}}_{\text{2}}\text{O}$, then ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is shown below.

Figure 4

Bromine and water oxidize the aldoses into carboxylic acids. The carboxylic acid thus formed is found in the form of lactones. Lactones formed are more stable in the five-membered $\text{γ-lactone}$ form.

Conclusion

The product obtained when D-mannose is reacted with ${\text{Br}}_{\text{2}}/{\text{H}}_{\text{2}}\text{O}$, then ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is shown in Figure 4.

Interpretation Introduction

(e)

Interpretation:

The products expected when D-mannose is reacted with ${\text{CH}}_{\text{3}}\text{OH},\text{HCl}$ is to be stated.

Concept introduction:

A monosaccharide is converted into cyclic acetals on reaction with alcohols in the presence of acidic conditions. The hydroxide group right to the oxygen atom in the pyranose ring structure is methylated and result in the formation of acetal.

Answer to Problem 24.34AP

The product obtained when D-mannose is reacted with ${\text{CH}}_{\text{3}}\text{OH},\text{HCl}$ is shown below.

Explanation of Solution

The product obtained when D-mannose is reacted with ${\text{CH}}_{\text{3}}\text{OH},\text{HCl}$ is shown below.

Figure 5

The D-mannose on reaction with methanol and hydrochloric acid is converted into the acetal. The acetal formed is found in both forms alpha and beta regardless of the configuration of D-mannose.

Conclusion

The product obtained when D-mannose is reacted with ${\text{CH}}_{\text{3}}\text{OH},\text{HCl}$ is shown in Figure 5.

Interpretation Introduction

(f)

Interpretation:

The products expected when D-mannose is reacted with acetic anhydride/pyridine is to be stated.

Concept introduction:

Esterification reaction of alcohol units of monosaccharide is also done. The esterification of the alcohol units is done in the presence of an excess of acetic anhydride and pyridine. All the alcohol units present are acylated.

Answer to Problem 24.34AP

The product obtained when D-mannose is reacted with acetic anhydride/pyridine is shown below.

Explanation of Solution

The product obtained when D-mannose is reacted with acetic anhydride/pyridine is shown below.

Figure 6

Esterification of all the alcohols units present in the D-mannose is done in the presence of acetic anhydride and pyridine.

Conclusion

The product obtained when D-mannose is reacted with acetic anhydride/pyridine is shown in Figure 6.

Interpretation Introduction

(g)

Interpretation:

The product obtained when the product of part (d) is reacted with $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$, then $\text{Fe}{\left(\text{OAc}\right)}_{\text{3}}$, ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ is to be stated.

Concept introduction:

Ruff’s degradation is the degradation of the sugar molecule by one carbon unit. The aldose is oxidized to the carboxylic acids using bromine water. The carboxylic acid is converted into calcium salt using the calcium hydroxide base which on further reaction with $\text{Fe}{\left(\text{OAc}\right)}_{\text{3}}$ and ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ degraded to the one-carbon less aldose.

Answer to Problem 24.34AP

The product obtained when the product of part (d) is reacted with $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$, then $\text{Fe}{\left(\text{OAc}\right)}_{\text{3}}$, ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ is shown below.

Explanation of Solution

The product of part (d) is shown below.

Figure 7

The product obtained when the product of part (d) is reacted with $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$, then $\text{Fe}{\left(\text{OAc}\right)}_{\text{3}}$, ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ is shown below.

Figure 8

The product obtained is the Ruff’s degradation product of the D-mannose. The D-mannonic acid is converted into calcium salt in the first step and then on reaction with $\text{Fe}{\left(\text{OAc}\right)}_{\text{3}}$ and ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ is converted into D-arabinose.

Conclusion

The product obtained when the product of part (d) is reacted with $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$, then $\text{Fe}{\left(\text{OAc}\right)}_{\text{3}}$, ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ is shown in Figure 8.

Interpretation Introduction

(h)

Interpretation:

The product obtained when the product of part (e) is reacted with ${\text{PhCH}}_{\text{2}}\text{Cl}$ (excess) and $\text{NaOH}$ is to be stated.

Concept introduction:

The alkylation of the hydroxyl group of sugars is an important reaction. The alkylation of hydroxyl groups is done with the help of alkylating agent alkyl halides in the presence of strong base sodium hydroxide.

Answer to Problem 24.34AP

The product obtained when the product of part (e) is reacted with ${\text{PhCH}}_{\text{2}}\text{Cl}$ (excess) and $\text{NaOH}$ is shown below.

Explanation of Solution

The product of part (e) is shown below.

Figure 9

The product obtained when the product of part (e) is reacted with ${\text{PhCH}}_{\text{2}}\text{Cl}$ (excess) and $\text{NaOH}$ is shown below.

Figure 10

The methyl D-mannopyranoside is alkylated in the strong base sodium hydroxide. The sodium hydroxide takes up the acidic proton of alcohol groups and converts them to alkoxide ion form. This alkoxide ion then substitutes the chloride group in the ${\text{PhCH}}_{\text{2}}\text{Cl}$ and give the alkylated product.

Conclusion

The product obtained when the product of part (e) is reacted with ${\text{PhCH}}_{\text{2}}\text{Cl}$ (excess) and $\text{NaOH}$ is shown in Figure 10.